# How to prove this property in a Poisson process?

For a Poisson process show, for $s < t$, that $P(N(s)=k|N(t)=n) = \binom{n}{k} (\frac{s}{t})^k (1-\frac{s}{t})^{n-k}$

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The joint probability mass function for $(N(s), N(t))$ for $s<t$ is, using the property of independent increments, i.e. $N(t) \sim N(s) + N(t-s)$: $$\mathbb{P}(N(s) = k, N(t) = n) = \mathbb{P}(N(s) = k, N(t) - N(s) = n - k) = \\\frac{(\mu s)^k}{k!} \mathrm{e}^{-\mu s} \cdot \frac{(\mu (t-s))^{n-k}}{(n-k)!} \mathrm{e}^{-\mu(t-s)} \cdot [ n-k \geqslant 0, k \geqslant 0]$$ Then, assuming $n \geqslant k \geqslant 0$: $$\mathbb{P}(N(s) =k | N(t) = n) = \frac{\mathbb{P}(N(s)=k,N(t)=n)}{\mathbb{P}(N(t)=n)} = \\ \frac{\frac{\mu^k s^k}{k!} \cdot \frac{\mu^{n-k} (t-s)^{n-k}}{(n-k)!} \cdot \mathrm{e}^{-\mu t} }{ \frac{\mu^n t^n}{n!} \mathrm{e}^{-\mu t}} = \left( \frac{s}{t} \right)^k \left( 1- \frac{s}{t} \right)^{n-k} \binom{n}{k}$$

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+1. Neat'n easy. –  Did Mar 16 '12 at 0:20

Let

$\ \ \ \ A$ be the number of events occurring in the interval $[0,s]$,

$\ \ \ \ B$ be the number of events occurring in the interval $[s,t]$,

and

$\ \ \ \ C$ be the number of events occurring in the interval $[0,t]$.

Using the fact that in a Poisson process, events occurring in one time frame are independent of events occurring in a different time frame, we may write the sought after probability as \tag{1} \eqalign{P ( A =k \mid C= n) &={P\bigl((A=k)\cap ( C=n) \bigr)\over P(C=n)}\cr &={P\bigl((A=k)\cap ( B=n-k) \bigr)\over P(C=n)}\cr &={P (A=k) P ( B=n-k) \over P(C=n)} .}

Now, if the Poisson process has parameter $\lambda$, then

$\ \ \ \ A$ has Poisson distribution with parameter $s\lambda$,

$\ \ \ \ B$ has Poisson distribution with parameter $(t-s)\lambda$,

and

$\ \ \ \ C$ has Poisson distribution with parameter $t\lambda$.

So, using equation $(1)$: \eqalign{ P( A =k \mid C= n) &={ {(s\lambda)^k e^{-s\lambda}\over k!} {( (t-s)\lambda)^{n-k} e^{-(t-s)\lambda}\over (n-k)!} \over {(t\lambda)^n e^{-t\lambda}\over n!}}\cr &={n\choose k} {s^k(t-s)^{n-k} \lambda^k \lambda^{n-k} e^{-t\lambda}\over t^n\lambda^ne^{-t\lambda}}\cr &={n\choose k} {s^k(t-s)^{n-k}\over t^kt^{n-k}}\cr &={n\choose k} {\Bigl({s\over t}\Bigr)^k \Bigl(1-{s\over t}\Bigr)^{n-k}}.\cr }

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